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BHU PMT BHU PMT (Screening) Solved Paper-2009

  • question_answer
    A car of mass m is moving with momentum p. If\[\mu \]be the coefficient of friction between the tyres and the road, what will be stopping distance due to friction alone?

    A)  \[\frac{{{p}^{2}}}{2\mu g}\]                      

    B)  \[\frac{{{p}^{2}}}{2m\mu g}\]

    C)   \[\frac{{{p}^{2}}}{2{{m}^{2}}\mu g}\]                  

    D)  \[\frac{{{p}^{2}}}{2mg}\]

    Correct Answer: C

    Solution :

                     If the car stops after covering distance 5, then \[F\times s=\frac{{{p}^{2}}}{2m}\] where                   F = friction Now      \[F=\mu mg\] So,          \[\mu mg\times s=\frac{{{p}^{2}}}{2m}\] \[\Rightarrow \]               \[s=\frac{{{p}^{2}}}{2\mu {{m}^{2}}g}\]


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